Question

Emissions of sulfur dioxide by industry sett off chemical changes in the atmosphere that result in "acid rain". The acidity of liquids is measured by pH on a scale of 0 to 14. Distilled water has pH 7.0, and lower pH values indicate acidity. Normal rain is somewhat acidic, so acid rain is sometimes defined as rainfall with a pH below 5.0. Suppose that pH measurements of rainfall on different days in a Canadian forest follow a Normal distribution with standard deviation σσ = 0.5. A sample of n days finds that the mean pH is x¯¯¯x¯ = 4.8. What are the P- Values for tests of sample sizes n = 5, n = 15 and n = 40?

n = 5:
n = 15:
n = 40:

Answer 1

a)

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (4.8 - 5)/(0.5/sqrt(5))
z = -0.89

P-value Approach
P-value = 0.1867

b)

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (4.8 - 5)/(0.5/sqrt(15))
z = -1.55

P-value Approach
P-value = 0.0606

c)

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (4.8 - 5)/(0.5/sqrt(40))
z = -2.53

P-value Approach
P-value = 0.0057